Lie's theorem


In mathematics, specifically the theory of Lie algebras, Lie's theorem states that,[1] over an algebraically closed field of characteristic zero, if is a finite-dimensional representation of a solvable Lie algebra, then there's a flag of invariant subspaces of with , meaning that for each and i.

Put in another way, the theorem says there is a basis for V such that all linear transformations in are represented by upper triangular matrices.[2] This is a generalization of the result of Frobenius that commuting matrices are simultaneously upper triangularizable, as commuting matrices generate an abelian Lie algebra, which is a fortiori solvable.

A consequence of Lie's theorem is that any finite dimensional solvable Lie algebra over a field of characteristic 0 has a nilpotent derived algebra (see #Consequences). Also, to each flag in a finite-dimensional vector space V, there correspond a Borel subalgebra (that consist of linear transformations stabilizing the flag); thus, the theorem says that is contained in some Borel subalgebra of .[1]



For algebraically closed fields of characteristic p>0 Lie's theorem holds provided the dimension of the representation is less than p (see the proof below), but can fail for representations of dimension p. An example is given by the 3-dimensional nilpotent Lie algebra spanned by 1, x, and d/dx acting on the p-dimensional vector space k[x]/(xp), which has no eigenvectors. Taking the semidirect product of this 3-dimensional Lie algebra by the p-dimensional representation (considered as an abelian Lie algebra) gives a solvable Lie algebra whose derived algebra is not nilpotent.



The proof is by induction on the dimension of   and consists of several steps. (Note: the structure of the proof is very similar to that for Engel's theorem.) The basic case is trivial and we assume the dimension of   is positive. We also assume V is not zero. For simplicity, we write  .

Step 1: Observe that the theorem is equivalent to the statement:[3]

  • There exists a vector in V that is an eigenvector for each linear transformation in  .

Indeed, the theorem says in particular that a nonzero vector spanning   is a common eigenvector for all the linear transformations in  . Conversely, if v is a common eigenvector, take   to its span and then   admits a common eigenvector in the quotient  ; repeat the argument.

Step 2: Find an ideal   of codimension one in  .

Let   be the derived algebra. Since   is solvable and has positive dimension,   and so the quotient   is a nonzero abelian Lie algebra, which certainly contains an ideal of codimension one and by the ideal correspondence, it corresponds to an ideal of codimension one in  .

Step 3: There exists some linear functional   in   such that


is nonzero. This follows from the inductive hypothesis (it is easy to check that the eigenvalues determine a linear functional).

Step 4:   is a  -invariant subspace. (Note this step proves a general fact and does not involve solvability.)

Let  ,  , then we need to prove  . If   then it's obvious, so assume   and set recursively  . Let   and   be the largest such that   are linearly independent. Then we'll prove that they generate U and thus   is a basis of U. Indeed, assume by contradiction that it's not the case and let   be the smallest such that  , then obviously  . Since   are linearly dependent,   is a linear combination of  . Applying the map   it follows that   is a linear combination of  . Since by the minimality of m each of these vectors is a linear combination of  , so is  , and we get the desired contradiction. We'll prove by induction that for every   and   there exist elements   of the base field such that   and


The   case is straightforward since  . Now assume that we have proved the claim for some   and all elements of   and let  . Since   is an ideal, it's  , and thus


and the induction step follows. This implies that for every   the subspace U is an invariant subspace of X and the matrix of the restricted map   in the basis   is upper triangular with diagonal elements equal to  , hence  . Applying this with   instead of X gives  . On the other hand, U is also obviously an invariant subspace of Y, and so


since commutators have zero trace, and thus  . Since   is invertible (because of the assumption on the characteristic of the base field),   and


and so  .

Step 5: Finish up the proof by finding a common eigenvector.

Write   where L is a one-dimensional vector subspace. Since the base field is algebraically closed, there exists an eigenvector in   for some (thus every) nonzero element of L. Since that vector is also eigenvector for each element of  , the proof is complete.  



The theorem applies in particular to the adjoint representation   of a (finite-dimensional) solvable Lie algebra   over an algebraically closed field of characteristic zero; thus, one can choose a basis on   with respect to which   consists of upper triangular matrices. It follows easily that for each  ,   has diagonal consisting of zeros; i.e.,   is a strictly upper triangular matrix. This implies that   is a nilpotent Lie algebra. Moreover, if the base field is not algebraically closed then solvability and nilpotency of a Lie algebra is unaffected by extending the base field to its algebraic closure. Hence, one concludes the statement (the other implication is obvious):[4]

A finite-dimensional Lie algebra   over a field of characteristic zero is solvable if and only if the derived algebra   is nilpotent.

Lie's theorem also establishes one direction in Cartan's criterion for solvability:

If V is a finite-dimensional vector space over a field of characteristic zero and   a Lie subalgebra, then   is solvable if and only if   for every   and  .[5]

Indeed, as above, after extending the base field, the implication   is seen easily. (The converse is more difficult to prove.)

Lie's theorem (for various V) is equivalent to the statement:[6]

For a solvable Lie algebra   over an algebraically closed field of characteristic zero, each finite-dimensional simple  -module (i.e., irreducible as a representation) has dimension one.

Indeed, Lie's theorem clearly implies this statement. Conversely, assume the statement is true. Given a finite-dimensional  -module V, let   be a maximal  -submodule (which exists by finiteness of the dimension). Then, by maximality,   is simple; thus, is one-dimensional. The induction now finishes the proof.

The statement says in particular that a finite-dimensional simple module over an abelian Lie algebra is one-dimensional; this fact remains true over any base field since in this case every vector subspace is a Lie subalgebra.[7]

Here is another quite useful application:[8]

Let   be a finite-dimensional Lie algebra over an algebraically closed field of characteristic zero with radical  . Then each finite-dimensional simple representation   is the tensor product of a simple representation of   with a one-dimensional representation of   (i.e., a linear functional vanishing on Lie brackets).

By Lie's theorem, we can find a linear functional   of   so that there is the weight space   of  . By Step 4 of the proof of Lie's theorem,   is also a  -module; so  . In particular, for each  ,  . Extend   to a linear functional on   that vanishes on  ;   is then a one-dimensional representation of  . Now,  . Since   coincides with   on  , we have that   is trivial on   and thus is the restriction of a (simple) representation of  .  

See also



  1. ^ a b Serre 2001, Theorem 3
  2. ^ Humphreys 1972, Ch. II, § 4.1., Corollary A.
  3. ^ Serre 2001, Theorem 3″
  4. ^ Humphreys 1972, Ch. II, § 4.1., Corollary C.
  5. ^ Serre 2001, Theorem 4
  6. ^ Serre 2001, Theorem 3'
  7. ^ Jacobson 1979, Ch. II, § 6, Lemma 5.
  8. ^ Fulton & Harris 1991, Proposition 9.17.


  • Fulton, William; Harris, Joe (1991). Representation theory. A first course. Graduate Texts in Mathematics, Readings in Mathematics. Vol. 129. New York: Springer-Verlag. doi:10.1007/978-1-4612-0979-9. ISBN 978-0-387-97495-8. MR 1153249. OCLC 246650103.
  • Humphreys, James E. (1972), Introduction to Lie Algebras and Representation Theory, Berlin, New York: Springer-Verlag, ISBN 978-0-387-90053-7.
  • Jacobson, Nathan (1979), Lie algebras (Republication of the 1962 original ed.), New York: Dover Publications, Inc., ISBN 0-486-63832-4, MR 0559927
  • Serre, Jean-Pierre (2001), Complex Semisimple Lie Algebras, Berlin: Springer, doi:10.1007/978-3-642-56884-8, ISBN 3-5406-7827-1, MR 1808366